Random Number Generator Detailed PPT with

Random Numbers 1
Random Number Generation
 Desirable Attributes:
– Uniformity
– Independence
– Efficiency
– Replicability
– Long Cycle Length

Random Numbers 2
(cont.)
Each random number Rt is an independent sample
drawn from a continuous uniform distribution
between 0 and 1
1 , 0 x 1
pdf: f(x) = 
0 , otherwise

Random Numbers 3
(cont.)
12
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1
4
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1
3
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1
)
2
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1
(
]
3
/
[
)]
(
[
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(
2
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1
]
2
/
[
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(
2
1
0
3
2
1
0
2
1
0
2
1
0












x
R
E
dx
x
R
V
x
dx
x
R
E
x
f(x)
0
1
PDF:

Random Numbers 4
Techniques for Generating
Random Number
MidSquare
Example:
X0 = 7182 (seed)
= 51581124
==> R1 = 0.5811
= (5811) 2
= 33767721
==> R2 = 0.7677
etc.
2
0
X
2
0
X

Random Numbers 5
Random Number (cont.)
Note: Cannot choose a seed that guarantees that the
sequence will not degenerate and will have a long
period. Also, zeros, once they appear, are carried in
subsequent numbers.
Ex1: X0 = 5197 (seed) = 27008809
==> R1 = 0.0088 = 00007744
==> R2 = 0.0077
Ex2: X0 = 4500 (seed) = 20250000
==> R1 = 0.2500 = 06250000
2
0
X
2
1
X
2
0
X
2
1
X

Random Numbers 6
 Multiplicative Congruential Method:
Basic Relationship
Xi+1 = a Xi (mod m), where a 0and m 0
Most natural choice for m is one that equals to the
capacity of a computer word.
m = 2b
(binary machine), where b is the number of
bits in the computer word.
m = 10d
(decimal machine), where d is the
number of digits in the computer word.

Random Numbers 7
The max period(P) is:
 For m a power of 2, say m = 2b
, and c 0, the longest
possible period is P = m = 2b
, which is achieved
provided that c is relatively prime to m (that is, the
greatest common factor of c and m is 1), and a = 1 +
4k, where k is an integer.
 For m a power of 2, say m = 2b
, and c =0, the longest
possible period is P = m / 4 = 2b-2
, which is achieved
provided that the seed X0 is odd and the multiplier, a, is
given by a = 3 + 8k or a = 5 + 8k, for some k = 0, 1,...

Random Numbers 8
For m a prime number and c =0, the longest
possible period is P = m - 1, which is achieved
provided that the multiplier, a, has the property
that the smallest integer k such that ak
- 1 is
divisible by m is k = m - 1,

Random Numbers 9
(Example)
Using the multiplicative congruential method, find
the period of the generator for a = 13, m = 26
, and
X0 = 1, 2, 3, and 4. The solution is given in next
slide. When the seed is 1 and 3, the sequence has
period 16. However, a period of length eight is
achieved when the seed is 2 and a period of length
four occurs when the seed is 4.

Random Numbers 10
Period Determination Using Various seeds
i Xi Xi Xi Xi
0 1 2 3 4
1 13 26 39 52
2 41 18 59 36
3 21 42 63 20
4 17 34 51 4
5 29 58 23
6 57 50 43
7 37 10 47
8 33 2 35
9 45 7
10 9 27
11 53 31
12 49 19
13 61 55
14 25 11
15 5 15
16 1 3

Random Numbers 11
SUBROUTINE RAN(IX, IY, RN)
IY = IX * 1220703125
IF (IY) 3,4,4
3: IY = IY + 214783647 + 1
4: RN = IY
RN = RN * 0.4656613E-9
IX = IY
RETURN
END

Random Numbers 12
 Linear Congruential Method:
Xi+1 = (aXi + c) mod m, i = 0, 1, 2....
(Example)
let X0 = 27, a = 17, c = 43, and m = 100, then
X1 = (17*27 + 43) mod 100 = 2
R1 = 2 / 100 = 0.02
X2 = (17*2 + 43) mod 100 = 77
R2 = 77 / 100 = 0.77
.........

Random Numbers 13
Test for Random Numbers
1. Frequency test. Uses the Kolmogorov-Smirnov or
the chi-square test to compare the distribution of the
set of numbers generated to a uniform distribution.
2. Runs test. Tests the runs up and down or the runs
above and below the mean by comparing the actual
values to expected values. The statistic for
comparison is the chi-square.
3. Autocorrelation test. Tests the correlation between
numbers and compares the sample correlation to the
expected correlation of zero.

Random Numbers 14
(cont.)
4. Gap test. Counts the number of digits that appear
between repetitions of a particular digit and then
uses the Kolmogorov-Smirnov test to compare
with the expected number of gaps.
5. Poker test. Treats numbers grouped together as a
poker hand. Then the hands obtained are
compared to what is expected using the chi-square
test.

Random Numbers 15
(cont.)
In testing for uniformity, the hypotheses are as
follows:
H0: Ri ~ U[0,1]
H1: Ri  U[0,1]
The null hypothesis, H0, reads that the numbers are
distributed uniformly on the interval [0,1].

Random Numbers 16
(cont.)
In testing for independence, the hypotheses are as
follows;
H0: Ri ~ independently
H1: Ri  independently
This null hypothesis, H0, reads that the numbers are
independent. Failure to reject the null hypothesis
means that no evidence of dependence has been
detected on the basis of this test. This does not imply
that further testing of the generator for independence
is unnecessary.

Random Numbers 17
(cont.)
Level of significance 
 = P(reject H0 | H0 true)
Frequently,  is set to 0.01 or 0.05
(Hypothesis)
Actually True Actually False
Accept 1 - 
(Type II error)
Reject  1 - 
(Type I error)

Random Numbers 18
(cont.)
 The Gap Test measures the number of digits between
successive occurrences of the same digit.
(Example) length of gaps associated with the digit 3.
4, 1, 3, 5, 1, 7, 2, 8, 2, 0, 7, 9, 1, 3, 5, 2, 7, 9, 4, 1, 6, 3
3, 9, 6, 3, 4, 8, 2, 3, 1, 9, 4, 4, 6, 8, 4, 1, 3, 8, 9, 5, 5, 7
3, 9, 5, 9, 8, 5, 3, 2, 2, 3, 7, 4, 7, 0, 3, 6, 3, 5, 9, 9, 5, 5
5, 0, 4, 6, 8, 0, 4, 7, 0, 3, 3, 0, 9, 5, 7, 9, 5, 1, 6, 6, 3, 8
8, 8, 9, 2, 9, 1, 8, 5, 4, 4, 5, 0, 2, 3, 9, 7, 1, 2, 0, 3, 6, 3
Note: eighteen 3’s in list
==> 17 gaps, the first gap is of length 10

Random Numbers 19
(cont.)
We are interested in the frequency of gaps.
P(gap of 10) = P(not 3) P(not 3) P(3) ,
note: there are 10 terms of the type P(not 3)
= (0.9)10
(0.1)
The theoretical frequency distribution for randomly
ordered digit is given by
F(x) = 0.1 (0.9)n
= 1 - 0.9x+1
Note: observed frequencies for all digits are
compared to the theoretical frequency using the
Kolmogorov-Smirnov test.
x
0
n


Random Numbers 20
(cont.)
(Example)
Based on the frequency with which gaps occur,
analyze the 110 digits above to test whether they are
independent. Use = 0.05. The number of gaps is
given by the number of digits minus 10, or 100. The
number of gaps associated with the various digits are
as follows:
Digit 0 1 2 3 4 5 6 7 8 9
# of Gaps 7 8 8 17 10 13 7 8 9 13

Random Numbers 21
(cont.)
Gap Test Example
Relative Cum. Relative
Gap Length Frequency Frequency Frequency F(x) |F(x) - SN(x)|
0-3 35 0.35 0.35 0.3439 0.0061
4-7 22 0.22 0.57 0.5695 0.0005
8-11 17 0.17 0.74 0.7176 0.0224
12-15 9 0.09 0.83 0.8147 0.0153
16-19 5 0.05 0.88 0.8784 0.0016
20-23 6 0.06 0.94 0.9202 0.0198
24-27 3 0.03 0.97 0.9497 0.0223
28-31 0 0.00 0.97 0.9657 0.0043
32-35 0 0.00 0.97 0.9775 0.0075
36-39 2 0.02 0.99 0.9852 0.0043
40-43 0 0.00 0.99 0.9903 0.0003
44-47 1 0.01 1.00 0.9936 0.0064

Random Numbers 22
(cont.)
The critical value of D is given by
D0.05 = 1.36 / 100 = 0.136
Since D = max |F(x) - SN(x)| = 0.0224 is less
than D0.05, do not reject the hypothesis of
independence on the basis of this test.

Random Numbers 23
(cont.)
 Run Tests (Up and Down)
Consider the 40 numbers; both the Kolmogorov-
Smirnov and Chi-square would indicate that the
numbers are uniformly distributed. But, not so.
0.08 0.09 0.23 0.29 0.42 0.55 0.58 0.72 0.89 0.91
0.11 0.16 0.18 0.31 0.41 0.53 0.71 0.73 0.74 0.84
0.02 0.09 0.30 0.32 0.45 0.47 0.69 0.74 0.91 0.95
0.12 0.13 0.29 0.36 0.38 0.54 0.68 0.86 0.88 0.91

Random Numbers 24
(cont.)
Now, rearrange and there is less reason to doubt
independence.
0.41 0.68 0.89 0.84 0.74 0.91 0.55 0.71 0.36 0.30
0.09 0.72 0.86 0.08 0.54 0.02 0.11 0.29 0.16 0.18
0.88 0.91 0.95 0.69 0.09 0.38 0.23 0.32 0.91 0.53
0.31 0.42 0.73 0.12 0.74 0.45 0.13 0.47 0.58 0.29

Random Numbers 25
(cont.)
Concerns:
 Number of runs
Length of runs
Note: If N is the number of numbers in a
sequence, the maximum number of runs is N-1,
and the minimum number of runs is one.
If “a” is the total number of runs in a sequence,
the mean and variance of “a” is given by

Random Numbers 26
(cont.)
a = (2n - 1) / 3
= (16N - 29) / 90
For N > 20, the distribution of “a” approximated
by a normal distribution, N(a , ).
This approximation can be used to test the
independence of numbers from a generator.
Z0= (a - a) / a
2
a

2
a


Random Numbers 27
Substituting for aanda ==>
Za = {a - [(2N-1)/3]} / {(16N-29)/90},
where Z ~ N(0,1)
Acceptance region for hypothesis of
independence -Z Z0  Z
(cont.)
/ 2
/ 2
-Z / 2
Z/ 2

Random Numbers 28
(cont.)
(Example)
Based on runs up and runs down, determine
whether the following sequence of 40 numbers is
such that the hypothesis of independence can be
rejected where  = 0.05.
0.41 0.68 0.89 0.94 0.74 0.91 0.55 0.62 0.36 0.27
0.19 0.72 0.75 0.08 0.54 0.02 0.01 0.36 0.16 0.28
0.18 0.01 0.95 0.69 0.18 0.47 0.23 0.32 0.82 0.53
0.31 0.42 0.73 0.04 0.83 0.45 0.13 0.57 0.63 0.29

Random Numbers 29
(cont.)
The sequence of runs up and down is as follows:
+ + + +  +  + +  + + + + + + + + + + + + 
There are 26 runs in this sequence. With N=40 and a=26,
a= {2(40) - 1} / 3 = 26.33 and
= {16(40) - 29} / 90 = 6.79
Then,
Z0 = (26 - 26.33) / 
Now, the critical value is Z0.025 = 1.96, so the
independence of the numbers cannot be rejected on the
basis of this test.
2
a


Random Numbers 30
(cont.)
 Poker Test - based on the frequency with which
certain digits are repeated.
Example:
0.255 0.577 0.331 0.414 0.828 0.909
Note: a pair of like digits appear in each number
generated.

Random Numbers 31
(cont.)
In 3-digit numbers, there are only 3 possibilities.
P(3 different digits) =
(2nd diff. from 1st) * P(3rd diff. from 1st & 2nd)
= (0.9) (0.8) = 0.72
P(3 like digits) =
(2nd digit same as 1st)  P(3rd digit same as 1st)
= (0.1) (0.1) = 0.01
P(exactly one pair) = 1 - 0.72 - 0.01 = 0.27

Random Numbers 32
(cont.)
(Example)
A sequence of 1000 three-digit numbers has been
generated and an analysis indicates that 680 have
three different digits, 289 contain exactly one pair
of like digits, and 31 contain three like digits.
Based on the poker test, are these numbers
independent?
Let  = 0.05.
The test is summarized in next table.

Random Numbers 33
(cont.)
Observed Expected (Oi - Ei)2
Combination, Frequency, Frequency, -----------
i Oi Ei Ei
Three different digits 680 720 2.24
Three like digits 31 10 44.10
Exactly one pair 289 270 1.33
1000 1000 47.65
The appropriate degrees of freedom are one less than
the number of class intervals. Since 2
0.05, 2 = 5.99 <
47.65, the independence of the numbers is rejected
on the basis of this test.

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